Question

1) Figure below show a 36 uL of stock solution containing 800 ppm of CuSO4 was transferred into T1 tube contained with 324 uL of wash buffer. Next, serial dilution was performed from T1 to T5. (CuSO4 molar mass: 159.61 g/mol) 36 ul 120 ML 120 ML 120 L 120 L 800 ppm Cuso T1 Stock solution T2 T3 T4 TS B1 Dilution Factor Final Concentration NA 800 ppm Tube Volume Diluent Total Final Transferred Volume Volume Volume (PL) (wash (After buffer) dilution) Stock NA NA NA NA T1 36 324 360 240 T2 120 240 360 240 T3 120 240 360 240 T4 120 360 240 T5 120 240 360 240 NA Not Available I. Fill in the blank a), b), c), d), and e). [5 marks] a b с d 240 e II. Convert 800 ppm to Molarity (M), ug/L, mg/L, ppb, and ppt. [5 marks] III. Calculate final concentration of T1 to T5 after being diluted. [5 marks] IV. If 100 PL of T5 solution are diluted to a final volume of 500 ML, what is the final concentration in Molarity of solution in B1? Convert the final calculation or the answer from M to uM. [2 marks]

Answer 1

1. Dilution factor:

To get the D.F., you divide the final volume by the initial volume.

Explanation: D.F.=V_f/V_i

a= 240/36= 6.67

b=c=d=e= 240/120 = 2

2. Molecular weight of CuSO4= 159.61 g/mol

1 ppm = 1 mg in in 1000 mL

800 ppm = 800 mg in 1000 mL

= 0.8 g in 1000 mL

=(0.8/159.61) moles in 1000 mL

= 0.0050 moles in 1000 mL

Now, molarity= number of moles in 1000 mL

Here, molarity= 0.0050 M

800 ppm= 800 mg/ Litre

=(800 x 1000 ) μg/ Litre

= 8 x10⁵ μg/ Litre

Amount in μg/ Litre is known as ppb.

So,ppb= 8 x10⁵

Now, 1 ppb= 1000 ppt

ppb means μg/ Litre and ppt means ng/Litre

1 μg= 1000 ng (nanogram)

So, ppt= 8 x10⁵ x 1000

= 8 x 10⁸

III. The equation needed to solve this question is as follows

• C₁ is the concentration of the stock solution.
• V₁ is the volume to be removed (i.e., aliquoted) from the concentrated stock solution.
• C₂ is the final concentration of the diluted solution.
• V₂ is the final volume of the diluted solution. This is the volume that results after V₁ from the stock solution has been diluted with diluent to achieve a total diluted volume of V₂.
• An alternative and commonly-used notation for this equation is M₁V₁ = M₂V₂, where M is used in place of C. Any kind of concentration can directly be replaced in place of C1 and C2. Here,we will use the ppm.

For T1, 36 x 800 = 360 x C2

or, C2= 80 ppm

For T2, 120 x 80 = 360 x C3

or, C3= 26.67 ppm

For T3, 120 x 26.67 = 360 x C4

or, C4= 8.89 ppm

For T4, 120 x 8.89 = 360 x C5

or, C5= 2.96 ppm

For T5, 129 x 2.96 = 360 x C6

or, C6= 0.99 ppm

IV. 100 x 0.99 = 500 x B1

or, B1= 0.2 ppm

1 ppm= 1mg/Litre

0.2 ppm = 0.2 mg/ Litre

= 0.0002 g/ Litre

=(0.0002/ 159.61) moles/ Litre

= 1.23 x 10^-3 moles/Litre

so, molarity= 1.23 x 10^-3 M

=1.238 x 10^-3 x 10⁶ μM

= 1238 μM